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The purpose of this lab is to acquaint you with using Maple to compute partial derivatives.
For a function of a single real variable, the derivative
gives information on whether the graph of is increasing or decreasing. Finding where the derivative is zero was important in finding extreme values. For a function of two (or more) variables, the situation is more complicated.
A differentiable function, , of two variables has two partial
. As you have learned in class, computing partial derivatives is
very much like computing regular derivatives. The main difference is
that when you are computing
, you must treat
the variable as if it was a constant and vice-versa when computing
The Maple commands for computing partial derivatives are D
and diff. The diff command can be used on both expressions and functions whereas the D command can be used only on functions. The examples below show all first order and second order partials in Maple.
> f := (x,y) -> x^2*y^2-x*y;
The next example shows how to evaluate the mixed partial derivative of the function given above at the point .
> der := diff(f(x,y),x,y);
The tangent plane like the tangent line to a single variable function is based on derivatives, however the partial derivatives are used for the tangent plane. Let's start with the equation of the tangent line to the function at the point where . Recall, the general equation of a line at the point having slope is
. This can be rewritten knowing that the derivative is the slope of a tangent line as
. Similarly for a funcion of two variables, the equation of the plane tangent to at the point has the equation
. The following example will show you how this can easily be translated to Maple syntax.
> g := x-> sin(x)-x^3/7+x^2;
> tanline := D(g)(5)*(x-5)+g(5);
The next example shows how to find the tangent plane to the function
at . You could write the partials with diff or D. This example uses D as it is easier to plug in the the point with this syntax; with diff the subs command would be used.
To find a point where the tangent plane is horizontal, you would need to solve where both first order partials are equal to zero simultaneously.
- Given the single variable function
- Plot over the interval
- Find the slope of at
- Find other values that have the same slope as the slope at
- Find the slope of the secant lines containing the value from part b and each value in part c.
- Find the equation of a single line that passes through both points from parts b and c and plot the function and the line on the same graph over the interval given in part a. Is the line tangent to the graph at both points?
- Find the tangent plane at .
- Plot the function and the tangent plane over the intervals
and rotate the plot so that you can see that the plane is tangent.
- There is only one point at which the plane tangent to the surface
is horizontal. Find it and plot it along with the function on the same graph. Be sure to use axes so that you can rotate the graph and see that the tangent plane is horizontal.
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Jane E Bouchard